Question

full answer of that question​

Answer 1

Step-by-step explanation:

Let the two circles with centre O and P intersect each other in points A and B, so AB is their common chord.

Therefore, OA = OB = 10 cm

And PA = PB = 8 cm.

AB = 12 cm.

$let \: OM \perp AB \: \\ and \: PM \perp AB \\ AM = \frac{1}{2} AB \\ = \frac{1}{2} \times 12 = 6 \: cm \\ in \: \triangle \: OAM \\ {OA}^{2} = {AM}^{2} + {OM}^{2} \\ {10}^{2} = {6}^{2} + {OM}^{2} \\ {OM}^{2} = 100 - 36 = 64 \\ OM = 8 \: cm \\ \\ next \: in \: \triangle \: PAM \\ {PA}^{2} = {AM}^{2} + {PM}^{2} \\ {8}^{2} = {6}^{2} + {PM}^{2} \\ {PM}^{2} = 64 - 36 = 28 \\ PM = \sqrt {28} = 5.29\: cm \\\therefore OP = OM + PM = 8 + 5.29= 13.29 \: cm. \\</p><p>\implies OP \approx 13.3 cm$

Therefore, the distance between the centres of the two circles is 13.3 cm.